Question

What isꝩfor a proton having a mass energy of 938.3MeV accelerated through an effective potential of 1.0TV (teravolt) at Fermilab outside Chicago?

Short answer

The proton is then obtained as ꝩ =1067.

Step-by-step solution

Identification of the given data

The mass energy of proton is 938.3MeV

Its effective potential is 1.0TV.

Define Relativistic Kinetic Energy

In essence, relativistic kinetic energy describes a particle's kinetic energy as the difference between that energy and its rest mass energy. This equation resembles the non-relativistic kinetic energy statement for low velocities.

Evaluating ꝩ for a proton

With the help of the equation (\[{\rm{28}}{\rm{.52}}\]), the relativistic kinetic energy is obtained by:

\[K{E_{rel}}=(\gamma-1)m{c^2}\] …(i)

We have the value of \[\gamma \] as the relativistic constant.

The value of \[m{c^2}\] is said to be the rest mass energy of the particle.

The rest mass of the proton is then obtained as:

\[m{c^2} = 938.3\,{\rm{MeV}}\].

The effective potential to accelerate the proton is then obtained as:

\[V = 1.0\,{\rm{TV}}\].

So, the kinetic energy of the proton is then evaluated as:

\begin{aligned}K{E_{rel{\rm{ }}}}={\rm{eV}}\\ &=1.0\,{\rm{TeV}}\\ &=1.0 \times {10^6}\,{\rm{MeV}}\end{aligned}

Now, with the help of the first equation, we get:

\begin{aligned}\gamma =\frac{{K{E_{rel}}}}{{m{c^2}}}+1\\ &=\frac{{1.0\times {{10}^6}{\rm{MeV}}}}{{938.3\,{\rm{MeV}}}}+1\\=1066.75\\ &= 1067\end{aligned}

Therefore, the proton is: \[\gamma = 1067\].